Lecture 2: Cm Types, Reflex Fields, and Main Theorem (2/22/12) Notes by I. Boreico

Recall the setting at the end of the previous lecture: A is an abelian variety over a field F of characteristic zero, with dimension g > 0. Let L be a number field of degree 2g over Q “acting” on A; that is, we have an embedding L ↪→ End(A). The elements of L that actually act on A – i.e., O := L ⋂ End(A) – form an order in L. Since char(F ) = 0, the action of O on T0(A) (a g-dimensional F -vector space) induces an action of L = O ⊗Z Q on T0(A). This makes T0(A) into an F ⊗Q L-module (also F ⊗Q L = F ⊗Z O, so we can get this action by another way). Keep in mind that there are two actions of Q (or Z) on Te(A): one coming from the embedding of Q in End(A) and another coming from the F -vector space structure (as A is an F -scheme). These actions coincide since both respect the underlying abelian group structure and the Z-module structure on an abelian group is unique (so likewise for a Q-vector space structure over this). If F = C, then A is a quotient V/Λ where V is a g-dimensional C-vector space and Λ is a O-module that is isomorphic to Z. In general when g > 1 this is not an invertible O-module when O 6 = OL. (The case g = 1 is very misleading in this regard since invertibility does hold in such cases for any O, due in part to the fact that quadratic orders are monogenic; higher-degree orders usually are not monogenic.) However, H1(A(C),Q) = ΛQ = Λ ⊗Z Q is a 1-dimensional L-vector space. We try as much as possible to work in the isogeny category (or to pass to the case when the CM order is OL) so that we can sidestep the fact that the integral homology is usually not an invertible O-module when O 6 = OL. Observe that T0(A) = V = R ⊗Z Λ ' R ⊗Q ΛQ as R ⊗Q L-modules, so it is a free R ⊗Q Lmodule of rank 1. This description (involving only R-linear structures) does not take into account the complex structure from F = C; we would like to know the structure of V = T0(A) as a module over C ⊗Q L ' ∏